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Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of a given triangle as shown in figure.

Now, (3,4) is the mid-point of AB, therefore,
{tex}3=\frac{x_1+x_2}{2}\ {/tex}and {tex}4=\frac{y_1+y_2}{2}{/tex}
x1 + x2 = 6 and y1 + y2 = 8 ..... (i)
(2,0) is the mid-point of BC, then,
{tex}2=\frac{x_2+x_3}{2}{/tex} and {tex} 0=\frac{y_2+y_3}{2}{/tex}
x2 + x3 = 4 and y2 + y3 = 0 ..........(ii)
(4,1) is the mid-point of AC, then,
{tex}4=\frac{x_1+x_3}{2}{/tex} and {tex}1=\frac{y_1+y_3}{2}{/tex}
x1 + x3 = 8 and y1 + y3 = 2 .........(iii)
Subtracting (ii) from (iii), we get,
x1 - x2 = 4 and y1 - y2 = 2 ........ (iv)
Adding (i) and (iv), we get,
2x1 = 10 and 2y1 = 10
x1 = 5 and y1 = 5
From (i), we have,
x2 = 6 - 5 = 1 and y2 = 8 - 5 = 3
From (ii), we have,
x3 = 4 - 1 = 3 and y3 = 0 - y2 = 0 - 3 = -3
Thus (5, 5), (1, 3) and (3, -3) are the vertices of triangle.
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Palak Sharma 4 years, 8 months ago
2Thank You